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bezimeni [28]
2 years ago
15

The discriminant of the equation x^2 - 4x + 4 = 0 shows there will be solution(s).

Mathematics
1 answer:
Alex787 [66]2 years ago
3 0

Answer:

one real

Answer: 1

Step-by-step explanation:

Givens

a = 1

b = -4

c = 4

Discriminate

Discriminate = sqrt(b^2 - 4*a*c)

Discriminate  = sqrt(  (-4)^2 - 4 * (1)(4) )

Discriminate = sqrt ( 16 - 16)

Discriminate = sqrt(0)

Discriminate = 0

That means the 1 real root is -  b / 2a = 4/(2*1) = - -2 = 2

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HCF of smallest 2 digit composite number and smallest 3 digit prime number is?​
posledela

Answer:

HCF = 1

Step-by-step explanation:

Smallest 2 digit composite number = 10

Smallest  3 digit prime number = 101

Factors of 10 = 1 , 2 , 5

Factors of 101 = 1 , 101

HCF = 1

6 0
2 years ago
X over 2x+1. What is x?
tatiyna

Answer:

x= - \frac{1}{2}

Step-by-step explanation:

Given:

\frac{2x+1}{x}

Let us assume that this equation is equal to 0.

Hence \frac{2x+1}{x}=0

Now Solving the above equation we get,

\frac{2x+1}{x}=0\\\\2x+1=0\\2x=-1\\x= - \frac{1}{2}

Hence by solving the equation we get x= - \frac{1}{2}

7 0
3 years ago
What is the square root of 85 and divide that by PI (3.14)​
stira [4]

√85 = 9.219544457

9.219544457 / π = 2.93467214

Round the answer as needed.

4 0
2 years ago
The nakagin capsule tower has 140 modules and is 14 stories high if the modules were divided evenly among the number of stories
ArbitrLikvidat [17]
To determine the number of modules that is in each story, we simply divide the number of modules by the number of stories. 
                           = 140 / 14
Simplifying,
                           = 10 modules / story
Therefore, by even distribution we note that there are 10 modules per story. 
8 0
3 years ago
Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 4 grams per liter is added at
Alla [95]

Answer:

a) dx/dt = 600 - 6x

b) x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F = 6 L/min

a) Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dx/dt = Fᵢxᵢ - Fx

Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min

dC/dt = 24 - 6C

dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25

(1/25)(dx/dt) = 24 - (6/25)x

dx/dt = 600 - 6x

b) dC/dt = 24 - 6C

dC/(24 - 6C) = dt

∫ dC/(24 - 6C) = ∫ dt

(-1/6) In (24 - C) = t + k (k = constant of integration)

In (24 - 6C) = -6t - 6k

-6k = K

In (24 - 6C) = K - 6t

At t = 0, C = 15 g/25 L = 0.6 g/L

In (24 - 6(0.6)) = K

In 20.4 = K

K = 3.02

So, the equation describing concentration of salt at anytime in the tank is

In (24 - 6C) = K - 6t

In (24 - 6C) = 3.02 - 6t

24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6

C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

But C = x/25

x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) when x = 20 g

20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

- (6t - 3.02) = In 80

- (6t - 3.02) = 4.382

(6t - 3.02) = -4.382

6t = -4.382 + 3.02

t = 1.362/6 = 0.227 min = 13.62 s

4 0
3 years ago
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